{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0046",
  "pilot_number": "P250-0096",
  "number": 96,
  "family_id": "qf-mod40-binomial-expansion",
  "title": {
    "zh-Hant": "第 96 題：二項式定理",
    "en": "Question 96: The Finite Expansion Includes Every k from Zero through n"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261109-r05-binomial-theorem-0046",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-binomial-theorem"
    ],
    "title_zh": "二項式定理",
    "stem_zh": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":20}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)",
      "explanation_zh": "Wave6第三層核對=(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 3,
      "skills": [
        "二項式定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095046,
      "parameters": {
        "n": 20,
        "domain": "Exact_commutative_binomial_theorem",
        "operation": "binomial-expansion",
        "semantic_template": "qf-mod40-binomial-expansion/{\"n\":20}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、single-element stabilizer、finite-prefix BW、single-cover compactness、zero-mass division、noncommutative binomial、incomplete Möbius factorization、bounded-search nonmembership、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0046",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "e0aedc127c9cafb8e44db0fe7320d1c84a2dad136d25e45209eefe470e64a538",
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095046,
      "semantic_template": "qf-mod40-binomial-expansion/{\"n\":20}"
    },
    "fields": {
      "title": "The Finite Expansion Includes Every k from Zero through n",
      "stem": "Use natural exponent n=20, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).",
      "answer": {
        "kind": "expression",
        "value": "(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)",
        "explanation": "Every row is [k,C(n,k),x exponent n−k,y exponent k]. Retain all n+1 rows and exact integer coefficients. At n=0 the expansion is constant1, including the formal zero-exponent convention. Noncommuting variables require a different theorem. The exact result, retaining the requested order, is (20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)."
      },
      "skills": [
        "The Finite Expansion Includes Every k from Zero through n"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":20}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use natural exponent n=20, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "n=20≥0; x,y commute",
      "after": "(x+y)^20 is a finite product of 20 factors",
      "relation": "implies",
      "rule_id": "finite_product",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個因子選 x 或 y，且乘法可交換，才能按 y 的選取次數合併同類項。",
        "en": "Choose x or y from each factor; commutativity permits grouping products by the number of y choices."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "此處不是非整數次方的無窮級數，也不套用到不可交換乘法。",
          "en": "This is not an infinite expansion for a noninteger power or a claim for noncommuting multiplication."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "choose exactly 0 of 20 factors to contribute y",
      "after": "C(20,0)=2432902008176640000/(1·2432902008176640000)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 0 個 y 位置與 20 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 0 y positions and 20 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "0"
        ],
        "output": "1",
        "label": "C(20,0)"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C(20,0)=1; choose y 0 times",
      "after": "1·x^20·y^0",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-0=20；指數相加仍為 20。",
        "en": "The x exponent is 20-0=20; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "choose exactly 1 of 20 factors to contribute y",
      "after": "C(20,1)=2432902008176640000/(1·121645100408832000)=20",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 1 個 y 位置與 19 個 x 位置各自的重排，得到係數 20。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 1 y positions and 19 x positions, giving coefficient 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "1"
        ],
        "output": "20",
        "label": "C(20,1)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "C(20,1)=20; choose y 1 times",
      "after": "20·x^19·y^1",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-1=19；指數相加仍為 20。",
        "en": "The x exponent is 20-1=19; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "choose exactly 2 of 20 factors to contribute y",
      "after": "C(20,2)=2432902008176640000/(2·6402373705728000)=190",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 2 個 y 位置與 18 個 x 位置各自的重排，得到係數 190。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 2 y positions and 18 x positions, giving coefficient 190."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "2"
        ],
        "output": "190",
        "label": "C(20,2)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "C(20,2)=190; choose y 2 times",
      "after": "190·x^18·y^2",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-2=18；指數相加仍為 20。",
        "en": "The x exponent is 20-2=18; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "choose exactly 3 of 20 factors to contribute y",
      "after": "C(20,3)=2432902008176640000/(6·355687428096000)=1140",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 3 個 y 位置與 17 個 x 位置各自的重排，得到係數 1140。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 3 y positions and 17 x positions, giving coefficient 1140."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "3"
        ],
        "output": "1140",
        "label": "C(20,3)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "C(20,3)=1140; choose y 3 times",
      "after": "1140·x^17·y^3",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-3=17；指數相加仍為 20。",
        "en": "The x exponent is 20-3=17; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "choose exactly 4 of 20 factors to contribute y",
      "after": "C(20,4)=2432902008176640000/(24·20922789888000)=4845",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 4 個 y 位置與 16 個 x 位置各自的重排，得到係數 4845。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 4 y positions and 16 x positions, giving coefficient 4845."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "4"
        ],
        "output": "4845",
        "label": "C(20,4)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "C(20,4)=4845; choose y 4 times",
      "after": "4845·x^16·y^4",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-4=16；指數相加仍為 20。",
        "en": "The x exponent is 20-4=16; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "choose exactly 5 of 20 factors to contribute y",
      "after": "C(20,5)=2432902008176640000/(120·1307674368000)=15504",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 5 個 y 位置與 15 個 x 位置各自的重排，得到係數 15504。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 5 y positions and 15 x positions, giving coefficient 15504."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "5"
        ],
        "output": "15504",
        "label": "C(20,5)"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "C(20,5)=15504; choose y 5 times",
      "after": "15504·x^15·y^5",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-5=15；指數相加仍為 20。",
        "en": "The x exponent is 20-5=15; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "choose exactly 6 of 20 factors to contribute y",
      "after": "C(20,6)=2432902008176640000/(720·87178291200)=38760",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 6 個 y 位置與 14 個 x 位置各自的重排，得到係數 38760。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 6 y positions and 14 x positions, giving coefficient 38760."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "6"
        ],
        "output": "38760",
        "label": "C(20,6)"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "C(20,6)=38760; choose y 6 times",
      "after": "38760·x^14·y^6",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-6=14；指數相加仍為 20。",
        "en": "The x exponent is 20-6=14; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "choose exactly 7 of 20 factors to contribute y",
      "after": "C(20,7)=2432902008176640000/(5040·6227020800)=77520",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 7 個 y 位置與 13 個 x 位置各自的重排，得到係數 77520。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 7 y positions and 13 x positions, giving coefficient 77520."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "7"
        ],
        "output": "77520",
        "label": "C(20,7)"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "C(20,7)=77520; choose y 7 times",
      "after": "77520·x^13·y^7",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-7=13；指數相加仍為 20。",
        "en": "The x exponent is 20-7=13; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "choose exactly 8 of 20 factors to contribute y",
      "after": "C(20,8)=2432902008176640000/(40320·479001600)=125970",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 8 個 y 位置與 12 個 x 位置各自的重排，得到係數 125970。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 8 y positions and 12 x positions, giving coefficient 125970."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "8"
        ],
        "output": "125970",
        "label": "C(20,8)"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "C(20,8)=125970; choose y 8 times",
      "after": "125970·x^12·y^8",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-8=12；指數相加仍為 20。",
        "en": "The x exponent is 20-8=12; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "choose exactly 9 of 20 factors to contribute y",
      "after": "C(20,9)=2432902008176640000/(362880·39916800)=167960",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 9 個 y 位置與 11 個 x 位置各自的重排，得到係數 167960。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 9 y positions and 11 x positions, giving coefficient 167960."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "9"
        ],
        "output": "167960",
        "label": "C(20,9)"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "C(20,9)=167960; choose y 9 times",
      "after": "167960·x^11·y^9",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-9=11；指數相加仍為 20。",
        "en": "The x exponent is 20-9=11; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "choose exactly 10 of 20 factors to contribute y",
      "after": "C(20,10)=2432902008176640000/(3628800·3628800)=184756",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 10 個 y 位置與 10 個 x 位置各自的重排，得到係數 184756。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 10 y positions and 10 x positions, giving coefficient 184756."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "10"
        ],
        "output": "184756",
        "label": "C(20,10)"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "C(20,10)=184756; choose y 10 times",
      "after": "184756·x^10·y^10",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-10=10；指數相加仍為 20。",
        "en": "The x exponent is 20-10=10; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "choose exactly 11 of 20 factors to contribute y",
      "after": "C(20,11)=2432902008176640000/(39916800·362880)=167960",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 11 個 y 位置與 9 個 x 位置各自的重排，得到係數 167960。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 11 y positions and 9 x positions, giving coefficient 167960."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "11"
        ],
        "output": "167960",
        "label": "C(20,11)"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "C(20,11)=167960; choose y 11 times",
      "after": "167960·x^9·y^11",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-11=9；指數相加仍為 20。",
        "en": "The x exponent is 20-11=9; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "choose exactly 12 of 20 factors to contribute y",
      "after": "C(20,12)=2432902008176640000/(479001600·40320)=125970",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 12 個 y 位置與 8 個 x 位置各自的重排，得到係數 125970。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 12 y positions and 8 x positions, giving coefficient 125970."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "12"
        ],
        "output": "125970",
        "label": "C(20,12)"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "C(20,12)=125970; choose y 12 times",
      "after": "125970·x^8·y^12",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-12=8；指數相加仍為 20。",
        "en": "The x exponent is 20-12=8; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "choose exactly 13 of 20 factors to contribute y",
      "after": "C(20,13)=2432902008176640000/(6227020800·5040)=77520",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 13 個 y 位置與 7 個 x 位置各自的重排，得到係數 77520。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 13 y positions and 7 x positions, giving coefficient 77520."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "13"
        ],
        "output": "77520",
        "label": "C(20,13)"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "C(20,13)=77520; choose y 13 times",
      "after": "77520·x^7·y^13",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-13=7；指數相加仍為 20。",
        "en": "The x exponent is 20-13=7; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S030",
      "uses": [
        "G001",
        "S029"
      ],
      "before": "choose exactly 14 of 20 factors to contribute y",
      "after": "C(20,14)=2432902008176640000/(87178291200·720)=38760",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 14 個 y 位置與 6 個 x 位置各自的重排，得到係數 38760。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 14 y positions and 6 x positions, giving coefficient 38760."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "14"
        ],
        "output": "38760",
        "label": "C(20,14)"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "C(20,14)=38760; choose y 14 times",
      "after": "38760·x^6·y^14",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-14=6；指數相加仍為 20。",
        "en": "The x exponent is 20-14=6; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S032",
      "uses": [
        "G001",
        "S031"
      ],
      "before": "choose exactly 15 of 20 factors to contribute y",
      "after": "C(20,15)=2432902008176640000/(1307674368000·120)=15504",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 15 個 y 位置與 5 個 x 位置各自的重排，得到係數 15504。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 15 y positions and 5 x positions, giving coefficient 15504."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "15"
        ],
        "output": "15504",
        "label": "C(20,15)"
      }
    },
    {
      "id": "S033",
      "uses": [
        "G001",
        "S032"
      ],
      "before": "C(20,15)=15504; choose y 15 times",
      "after": "15504·x^5·y^15",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-15=5；指數相加仍為 20。",
        "en": "The x exponent is 20-15=5; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S034",
      "uses": [
        "G001",
        "S033"
      ],
      "before": "choose exactly 16 of 20 factors to contribute y",
      "after": "C(20,16)=2432902008176640000/(20922789888000·24)=4845",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 16 個 y 位置與 4 個 x 位置各自的重排，得到係數 4845。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 16 y positions and 4 x positions, giving coefficient 4845."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "16"
        ],
        "output": "4845",
        "label": "C(20,16)"
      }
    },
    {
      "id": "S035",
      "uses": [
        "G001",
        "S034"
      ],
      "before": "C(20,16)=4845; choose y 16 times",
      "after": "4845·x^4·y^16",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-16=4；指數相加仍為 20。",
        "en": "The x exponent is 20-16=4; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S036",
      "uses": [
        "G001",
        "S035"
      ],
      "before": "choose exactly 17 of 20 factors to contribute y",
      "after": "C(20,17)=2432902008176640000/(355687428096000·6)=1140",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 17 個 y 位置與 3 個 x 位置各自的重排，得到係數 1140。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 17 y positions and 3 x positions, giving coefficient 1140."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "17"
        ],
        "output": "1140",
        "label": "C(20,17)"
      }
    },
    {
      "id": "S037",
      "uses": [
        "G001",
        "S036"
      ],
      "before": "C(20,17)=1140; choose y 17 times",
      "after": "1140·x^3·y^17",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-17=3；指數相加仍為 20。",
        "en": "The x exponent is 20-17=3; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S038",
      "uses": [
        "G001",
        "S037"
      ],
      "before": "choose exactly 18 of 20 factors to contribute y",
      "after": "C(20,18)=2432902008176640000/(6402373705728000·2)=190",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 18 個 y 位置與 2 個 x 位置各自的重排，得到係數 190。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 18 y positions and 2 x positions, giving coefficient 190."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "18"
        ],
        "output": "190",
        "label": "C(20,18)"
      }
    },
    {
      "id": "S039",
      "uses": [
        "G001",
        "S038"
      ],
      "before": "C(20,18)=190; choose y 18 times",
      "after": "190·x^2·y^18",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-18=2；指數相加仍為 20。",
        "en": "The x exponent is 20-18=2; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S040",
      "uses": [
        "G001",
        "S039"
      ],
      "before": "choose exactly 19 of 20 factors to contribute y",
      "after": "C(20,19)=2432902008176640000/(121645100408832000·1)=20",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 19 個 y 位置與 1 個 x 位置各自的重排，得到係數 20。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 19 y positions and 1 x positions, giving coefficient 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "19"
        ],
        "output": "20",
        "label": "C(20,19)"
      }
    },
    {
      "id": "S041",
      "uses": [
        "G001",
        "S040"
      ],
      "before": "C(20,19)=20; choose y 19 times",
      "after": "20·x^1·y^19",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-19=1；指數相加仍為 20。",
        "en": "The x exponent is 20-19=1; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S042",
      "uses": [
        "G001",
        "S041"
      ],
      "before": "choose exactly 20 of 20 factors to contribute y",
      "after": "C(20,20)=2432902008176640000/(2432902008176640000·1)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 20 個位置，再除掉 20 個 y 位置與 0 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 20 positions and divide by reorderings within the 20 y positions and 0 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "20",
          "20"
        ],
        "output": "1",
        "label": "C(20,20)"
      }
    },
    {
      "id": "S043",
      "uses": [
        "G001",
        "S042"
      ],
      "before": "C(20,20)=1; choose y 20 times",
      "after": "1·x^0·y^20",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 20-20=0；指數相加仍為 20。",
        "en": "The x exponent is 20-20=0; the two exponents still sum to 20."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S044",
      "uses": [
        "G001",
        "S043"
      ],
      "before": "all k=0 through 20 have been accounted for",
      "after": "(x+y)^20=1·x^20·y^0 + 20·x^19·y^1 + 190·x^18·y^2 + 1140·x^17·y^3 + 4845·x^16·y^4 + 15504·x^15·y^5 + 38760·x^14·y^6 + 77520·x^13·y^7 + 125970·x^12·y^8 + 167960·x^11·y^9 + 184756·x^10·y^10 + 167960·x^9·y^11 + 125970·x^8·y^12 + 77520·x^7·y^13 + 38760·x^6·y^14 + 15504·x^5·y^15 + 4845·x^4·y^16 + 1140·x^3·y^17 + 190·x^2·y^18 + 20·x^1·y^19 + 1·x^0·y^20",
      "relation": "implies",
      "rule_id": "collect_all_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一種 x/y 選取都落入唯一的 k 類，因此這是完整有限展開，不是只列頭尾項。",
        "en": "Every x/y choice belongs to exactly one k class, so this is the complete finite expansion, not just endpoint terms."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "共有 n+1 個係數位置，k 的全範圍逐一保留。",
          "en": "There are n+1 coefficient positions, with the complete range of k retained."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "n": 20,
    "terms": [
      [
        0,
        1,
        20,
        0
      ],
      [
        1,
        20,
        19,
        1
      ],
      [
        2,
        190,
        18,
        2
      ],
      [
        3,
        1140,
        17,
        3
      ],
      [
        4,
        4845,
        16,
        4
      ],
      [
        5,
        15504,
        15,
        5
      ],
      [
        6,
        38760,
        14,
        6
      ],
      [
        7,
        77520,
        13,
        7
      ],
      [
        8,
        125970,
        12,
        8
      ],
      [
        9,
        167960,
        11,
        9
      ],
      [
        10,
        184756,
        10,
        10
      ],
      [
        11,
        167960,
        9,
        11
      ],
      [
        12,
        125970,
        8,
        12
      ],
      [
        13,
        77520,
        7,
        13
      ],
      [
        14,
        38760,
        6,
        14
      ],
      [
        15,
        15504,
        5,
        15
      ],
      [
        16,
        4845,
        4,
        16
      ],
      [
        17,
        1140,
        3,
        17
      ],
      [
        18,
        190,
        2,
        18
      ],
      [
        19,
        20,
        1,
        19
      ],
      [
        20,
        1,
        0,
        20
      ]
    ],
    "boundary": "complete_k_zero_through_n_coefficients"
  },
  "conclusion": {
    "canonical_value": "(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)",
    "explanation": {
      "zh-Hant": "完整展開如上述所有單項式；欄序表逐項對應。",
      "en": "The full expansion consists of all displayed monomials, in the same order as the answer rows."
    }
  },
  "definition_note": {
    "zh-Hant": "n 是非負整數，x 與 y 可交換；n=0 使用多項式空乘積單位元。",
    "en": "n is a nonnegative integer and x,y commute. For n=0 use the polynomial empty-product identity."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
      "batch_path": "public/data/questions/batches/batch-20261014-advance-20261109-r05-binomial-theorem-foundations/questions.jsonl",
      "batch_sha256": "f83302c1353681ecb1209ee1ccbef0d266f15fb9415191ed9289eb329e1175cc",
      "original_line": 46,
      "original_line_sha256": "13b13d4534844d476b4c4b31100e5eaf1295b4962287185d330a47215457e497",
      "question_compact_sha256": "e0aedc127c9cafb8e44db0fe7320d1c84a2dad136d25e45209eefe470e64a538",
      "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261109-r05-binomial-theorem-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261109-r05-binomial-theorem-foundations.en.jsonl",
      "english_row_compact_sha256": "7946207968d55b3e2a5c0eb0593f69d070f68298cc4eb21192d1440c0f85578f"
    },
    "mko_bindings": {
      "mko-binomial-theorem": {
        "sha256": "c8fefe3783fc4f7633b4b22b8cf149d96a683900f181e3b42c690409ce90e5b7",
        "english_sha256": "1086dfe7c31f4e6bf8934681a0a080b50c14cb828f1a8b17e59143ec2cf06c83",
        "original_path": "public/data/mko/binomial-theorem.json",
        "english_path": "public/data/i18n/editions/mko-binomial-theorem.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "fb325b42cfcaa25ddbae992459c2684afd767fe218028bc0b9c0823e67c83b7c",
    "original_candidate_sha256": "aaa73afe670f64de9f7accbd0b3d46b930e87628221b6f52633436f7658aa8f9",
    "original_candidate_line_sha256": "aaa73afe670f64de9f7accbd0b3d46b930e87628221b6f52633436f7658aa8f9"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "本稿是來源綁定的教學推導候選；精確重算不等於獨立數學審查或形式化證明。原題、答案、公式、AST 與歷史未被修改。",
      "en": "This is a source-bound pedagogical derivation candidate. Exact recomputation is not independent mathematical review or formal proof. Original questions, answers, formulas, ASTs and history are unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261024-002",
    "review_date": "2026-10-24",
    "owner_authorized_reader_publication": true,
    "scope": "Family prototypes and mathematical/bilingual reasoning variants reviewed in this existing context; parameter substitutions, all original/source bindings, full answer tuples and arithmetic ASTs independently reconstructed per question. Numeric binomial text slots are validated against independently regenerated n,k,C(n,k),n-k values. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "4727d1bd957a9b7f6cadb7972ddb5ab80a31f5259dc5394c1dff2b71944ed36b",
      "question_first_witness_sha256": "51e18649133238ac992229f6e16542a5f3af53ca73b2262ab0524508ea937dee"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 44,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "c70f67cfbb3513be4d87edd7e3f600edf75c15972265f23104b3e9f1dd2ab9d3",
      "arithmetic_steps": 21
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}